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  "body": "\n## Definition\n\nLet $n \\ge 2$ and let $\\Omega \\subseteq \\R^n$ be a bounded [[open set|02BH3MCBQB]] whose [[boundary|05D5EGMFYV]] $\\partial\\Omega$ is a $C^1$ hypersurface.\nLet $\\sigma$ be surface measure on $\\partial\\Omega$ and $\\nu(y)$ the outward unit normal at $y \\in \\partial\\Omega$.\nLet $\\Phi$ be the [[fundamental solution of the Laplacian|BSJ8X2XRN9]], normalised by $-\\Delta\\Phi = \\delta_0$.\nLet $\\varphi \\colon \\partial\\Omega \\to \\R$ be continuous.\nThe *double layer potential* with density $\\varphi$ is the function\n$$D\\varphi(x) = \\int_{\\partial\\Omega} \\frac{\\partial \\Phi(x - y)}{\\partial \\nu(y)}\\, \\varphi(y)\\, d\\sigma(y), \\qquad x \\in \\R^n \\setminus \\partial\\Omega,$$\nwhere $\\dfrac{\\partial \\Phi(x - y)}{\\partial \\nu(y)} = \\nu(y) \\cdot \\nabla_y \\big[\\Phi(x - y)\\big]$ is the derivative in $y$ along the outward normal.\n\n## Remarks\n\nThe integral is a [[surface integral|065HE5V98E]] with an integrand that is smooth in $x$ off $\\partial\\Omega$, and $D\\varphi$ is [[harmonic|0DEB0JMN0W]] on $\\R^n \\setminus \\partial\\Omega$.\nFor $n = 3$, $\\Phi(x) = 1/(4\\pi\\abs{x})$ and the kernel is $\\dfrac{\\nu(y) \\cdot (x - y)}{4\\pi \\abs{x - y}^3}$.\n\nExample (Gauss's integral): let $\\varphi \\equiv 1$.\nBy the [[divergence theorem|0JX86PSSA9]], $D1(x) = -1$ for $x \\in \\Omega$ and $D1(x) = 0$ for $x \\notin \\overline{\\Omega}$.\nFor the unit ball in $\\R^3$ and $x = 0$ this is a direct check: $\\nu(y) = y$, so the kernel is $y \\cdot (-y)/(4\\pi) = -1/(4\\pi)$, and the sphere has area $4\\pi$.\n\nNon-example: replacing the normal derivative of $\\Phi$ by $\\Phi$ itself gives the [[single layer potential|9T6ZEV4125]] instead.\nFor the unit ball in $\\R^3$ with $\\varphi \\equiv 1$, that potential equals $1$ inside the ball, while the double layer potential equals $-1$ there.\n",
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